Answer
$v_0 = 23~ft/s$
Work Step by Step
We can find an expression for $v_x$:
$tan~\theta = \frac{v_{0y}}{v_x}$
$v_x = \frac{v_{0y}}{tan~\theta}$
We can find an expression for the time of flight:
$x = v_x~t$
$t = \frac{x}{v_x}$
$t = \frac{x}{\frac{v_{0y}}{tan~\theta}}$
$t = \frac{x~tan~\theta}{v_{0y}}$
We can find $v_{0y}$:
$y = y_0+v_{0y}~t+\frac{1}{2}a_y~t^2$
$y = y_0+v_{0y}~\frac{x~tan~\theta}{v_{0y}}+\frac{1}{2}(a_y)~(\frac{x~tan~\theta}{v_{0y}})^2$
$y - y_0-x~tan~\theta = \frac{a_y~(x~tan~\theta)^2}{2~v_{0y}^2}$
$v_{0y}^2 = \frac{a_y~(x~tan~\theta)^2}{2~(y - y_0-x~tan~\theta)}$
$v_{0y} = \sqrt{\frac{a_y~(x~tan~\theta)^2}{2~(y - y_0-x~tan~\theta)}}$
$v_{0y} = \sqrt{\frac{(-32~ft/s^2)~(13~ft)^2~(tan~55^{\circ})^2}{2~(10~ft - 7.0~ft-(13~ft)~tan~55^{\circ})}}$
$v_{0y} = 18.82~ft/s$
We can find $v_0$:
$sin~\theta = \frac{v_{0y}}{v_0}$
$v_0 = \frac{v_{0y}}{sin~\theta}$
$v_0 = \frac{18.82~ft/s}{sin~55^{\circ}}$
$v_0 = 23~ft/s$
